This is a quadratic equation, with a vertex at the top.
#(x+2)^2 = -8(y-1)# --> Expand the left hand side
#x^2+4x+4 = -8(y-1)# --> Divide both sides by -1/8
#-1/8x^2-1/2x-1/2= y-1# --> Add one to both sides
#-1/8x^2-1/2x-1/2+1= y# --> Simplify
#y=-1/8x^2-1/2x+1/2#
Apply the equation for the #x#-value of the vertex of a parabola. For a quadratic of the form #y=ax^2+bx+c#, the #x#-value of the vertex is
#x=-b/(2a)#
For #b=-1/2# and #a=-1/8#, we have vertex #x=-2#. Plugging #x=-2# into #y=-1/8(-2)^2-1/2(-2)+1/2# gives #y=1#. So the vertex is #(-2,1)#.
Graphically,
graph{-1/8x^2-.5x+.5 [-10, 10, -5, 5]}
